Additive map

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In algebra, an additive map or additive function is a function ${\displaystyle f}$ that preserves the addition operation:^[1] ${\displaystyle f(x+y)=f(x)+f(y)}$ for every pair of elements ${\displaystyle x}$ and ${\displaystyle y}$ in the domain of ⁠ ${\displaystyle f}$ ⁠. For example, any linear map is additive. When the domain is the integers (ℤ), this is called a ${\displaystyle \mathbb {Z} }$ -linear map. When the domain is the real numbers, this is Cauchy's functional equation. For a specific case of this definition, see additive polynomial.

More formally, an additive map is a ${\displaystyle \mathbb {Z} }$ -module homomorphism. Since an abelian group is a ${\displaystyle \mathbb {Z} }$ -module, it may be defined as a group homomorphism between abelian groups.

A map ${\displaystyle V\times W\to X}$ that is additive in each of two arguments separately is called a bi-additive map or a ${\displaystyle \mathbb {Z} }$ -bilinear map.^[2]

Examples#edit

Typical examples include maps between rings, vector spaces, or modules that preserve the additive group. An additive map does not necessarily preserve any other structure of the object; for example, the product operation of a ring.

If ${\displaystyle f}$ and ${\displaystyle g}$ are additive maps, then the map ${\displaystyle f+g}$ (defined pointwise) is additive.

Properties#edit

Definition of scalar multiplication by an integer

Suppose that ${\displaystyle X}$ is an additive group with identity element ${\displaystyle 0}$ and that the inverse of ${\displaystyle x\in X}$ is denoted by ⁠ ${\displaystyle -x}$ ⁠. For any ${\displaystyle x\in X}$ and integer ⁠ ${\displaystyle n\in \mathbb {Z} }$ ⁠, let: ${\displaystyle nx:=\left\\begin{alignedat}{9}&&&0&&&&&&~~~~&&&&~{\text{ when n=0\\&&&x&&+\cdots +&&x&&~~~~{\text{(n&&{\text{ summands) &&~{\text{ when n>0\\&(-&&x)&&+\cdots +(-&&x)&&~~~~{\text{(|n&$ Thus ${\displaystyle (-1)x=-x}$ and it can be shown that for all integers ${\displaystyle m,n\in \mathbb {Z} }$ and all ⁠ ${\displaystyle x\in X}$ ⁠, ${\displaystyle (m+n)x=mx+nx}$ and ⁠ ${\displaystyle -(nx)=(-n)x=n(-x)}$ ⁠. This definition of scalar multiplication makes the cyclic subgroup ${\displaystyle \mathbb {Z} x}$ of ${\displaystyle X}$ into a left ${\displaystyle \mathbb {Z} }$ -module; if ${\displaystyle X}$ is commutative, then it also makes ${\displaystyle X}$ into a left ${\displaystyle \mathbb {Z} }$ -module.

Homogeneity over the integers

If ${\displaystyle f:X\to Y}$ is an additive map between additive groups then ${\displaystyle f(0)=0}$ and for all ⁠ ${\displaystyle x\in X}$ ⁠, ${\displaystyle f(-x)=-f(x)}$ (where negation denotes the additive inverse) and^[3] ${\displaystyle f(nx)=nf(x)\quad {\text{ for all n\in \mathbb {Z} .}$ Consequently, ${\displaystyle f(x-y)=f(x)-f(y)}$ for all ${\displaystyle x,y\in X}$ (where, by definition, ⁠ ${\displaystyle x-y:=x+(-y)}$ ⁠).

In other words, every additive map is homogeneous over the integers. Consequently, every additive map between abelian groups is a homomorphism of ${\displaystyle \mathbb {Z} }$ -modules.

Homomorphism of ${\displaystyle \mathbb {Q} }$ -modules

If the additive abelian groups ${\displaystyle X}$ and ${\displaystyle Y}$ are also a unital modules over the rationals ${\displaystyle \mathbb {Q} }$ (such as real or complex vector spaces) then an additive map ${\displaystyle f:X\to Y}$ satisfies:^[4] ${\displaystyle f(qx)=qf(x)\quad {\text{ for all q\in \mathbb {Q} {\text{ and x\in X.}$ In other words, every additive map is homogeneous over the rational numbers. Consequently, every additive maps between unital ${\displaystyle \mathbb {Q} }$ -modules is a homomorphism of ${\displaystyle \mathbb {Q} }$ -modules.

Despite being homogeneous over ⁠ ${\displaystyle \mathbb {Q} }$ ⁠, as described in the article on Cauchy's functional equation, even when ⁠ ${\displaystyle X=Y=\mathbb {R} }$ ⁠, it is nevertheless still possible for the additive function ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ to not be homogeneous over the real numbers; said differently, there exist additive maps ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ that are not of the form ${\displaystyle f(x)=s_{0}x}$ for some constant ⁠ ${\displaystyle s_{0}\in \mathbb {R} }$ ⁠. In particular, there exist additive maps that are not linear maps with respect to an existing ring structure of the codomain.

Notes#edit

^ Leslie Hogben (2013), Handbook of Linear Algebra (3 ed.), CRC Press, p. 30–8, ISBN 9781498785600
^ N. Bourbaki (1989), Algebra Chapters 1–3, Springer, p. 243
^ ${\displaystyle f(0)=f(0+0)=f(0)+f(0)}$ so adding ${\displaystyle -f(0)}$ to both sides proves that ⁠ ${\displaystyle f(0)=0}$ ⁠. If ${\displaystyle x\in X}$ then ${\displaystyle 0=f(0)=f(x+(-x))=f(x)+f(-x)}$ so that ${\displaystyle f(-x)=-f(x)}$ where, by definition, ⁠ ${\displaystyle (-1)f(x):=-f(x)}$ ⁠. Induction shows that if ${\displaystyle n\in \mathbb {N} }$ is positive then ${\displaystyle f(nx)=nf(x)}$ and that the additive inverse of ${\displaystyle nf(x)}$ is ⁠ ${\displaystyle n(-f(x))}$ ⁠, which implies that ${\displaystyle f((-n)x)=f(n(-x))=nf(-x)=n(-f(x))=-(nf(x))=(-n)f(x)}$ (this shows that ${\displaystyle f(nx)=nf(x)}$ holds for ⁠ ${\displaystyle n<0}$ ⁠). ${\displaystyle \blacksquare }$
^ Let ${\displaystyle x\in X}$ and ${\displaystyle q={\frac {m}{n\in \mathbb {Q} }$ where ${\displaystyle m,n\in \mathbb {Z} }$ and ⁠ ${\displaystyle n>0}$ ⁠. Let ⁠ ${\displaystyle y:={\frac {1}{nx}$ ⁠. Then ⁠ ${\displaystyle ny=n\left({\frac {1}{nx\right)=\left(n{\frac {1}{n\right)x=(1)x=x}$ ⁠, which implies ${\displaystyle f(x)=f(ny)=nf(y)=nf\left({\frac {1}{nx\right)}$ so that multiplying both sides by ${\displaystyle {\frac {1}{n$ proves that ⁠ ${\displaystyle f\left({\frac {1}{nx\right)={\frac {1}{nf(x)}$ ⁠. Consequently, ⁠ ${\displaystyle f(qx)=f\left({\frac {m}{nx\right)=mf\left({\frac {1}{nx\right)=m\left({\frac {1}{nf(x)\right)=qf(x)}$ ⁠. ${\displaystyle \blacksquare }$

Proofs

References#edit

Roger C. Lyndon; Paul E. Schupp (2001), Combinatorial Group Theory, Springer

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Additive map

Examples#edit

Properties#edit

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References#edit